Four sources of sound each of sound level 10, | vedclass

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(C) The sound level in decibels $(dB)$ is given by the formula: $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$.Given that each source has a sound lev

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$16$

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(C) The sound level in decibels $(dB)$ is given by the formula: $L = 10 \log_{10} \left( \frac{I}{I_0} ight)$.Given that each source has a sound level of $10\,dB$,we have $10 = 10 \log_{10} \left( \frac{I}{I_0} ight)$,which implies $\log_{10} \left( \frac{I}{I_0} ight) = 1$,so $I = 10 I_0$.When four such sources are sounded together,the total intensity $I_{	ext{total}} = 4 	imes I = 4 	imes 10 I_0 = 40 I_0$.The resultant sound level $L_{	ext{total}}$ is $10 \log_{10} \left( \frac{40 I_0}{I_0} ight) = 10 \log_{10}(40)$.Since $\log_{10}(40) = \log_{10}(4 	imes 10) = \log_{10}(4) + \log_{10}(10) \approx 0.602 + 1 = 1.602$.Therefore,$L_{	ext{total}} = 10 	imes 1.602 = 16.02\,dB \approx 16\,dB$.

Four sources of sound each of sound level $10\,dB$ are sounded together,the resultant intensity level will be ... $dB$

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